数组中占比超过一半的元素称之为主要元素。给你一个 整数 数组，找出其中的主要元素。若没有，返回 -1 。请设计时间复杂度为 O(N) 、空间复杂度为 O(1) 的解决方案。

难度：简单

示例 1：

输入：[1,2,5,9,5,9,5,5,5]
输出：5
示例 2：

输入：[3,2]
输出：-1
示例 3：

输入：[2,2,1,1,1,2,2]
输出：2

链接：https://leetcode-cn.com/problems/find-majority-element-lcci

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.Arrays;

class Solution {
    public int majorityElement(int[] nums) {
        int candidate = -1;
        int count = 0;
        for (int num : nums) {
            if (count == 0) {
                candidate = num;
            }
            if (num == candidate) {
                count++;
            } else {
                count--;
            }
        }
        count = 0;
        int length = nums.length;
        for (int num : nums) {
            if (num == candidate) {
                count++;
            }
        }
        return count * 2 > length ? candidate : -1;
    }

    public int[] stringToInt(String[] arr) {
        int[] ints = new int[arr.length];
        for (int i=0; i<arr.length; i++) {
            ints[i] = Integer.parseInt(arr[i]);
        }
        return ints;
    }

    public static void main(String[] args) throws IOException {
        // int[] nums = {1,2,5,9,5,9,5,5,5};
        // System.out.println(new Solution().majorityElement(nums));
        BufferedReader br =
                new BufferedReader(new InputStreamReader(System.in));
        String str;
        while ((str = br.readLine()) != null) {
            try {
                // nums
                String[] numsArray =
                        str.replace("[", "").replace("]", "").split(",");
                System.out.println("numsArray: " + Arrays.toString(numsArray));

                int[] nums = new Solution().stringToInt(numsArray);
                System.out.println(new Solution().majorityElement(nums));
            } catch (Exception e) {
                // e.printStackTrace();
                // 回车一行就退出
                break;
            }
        }
    }
}

View Code

 

连续输入[1,2,5,9,5,9,5,5,5]:

public static void main(String[] args) throws IOException {
    // int[] nums = {1,2,5,9,5,9,5,5,5};
    // System.out.println(new Solution().majorityElement(nums));
    BufferedReader br =
            new BufferedReader(new InputStreamReader(System.in));
    String str;
    while ((str = br.readLine()) != null) {
        try {
            // nums
            String[] numsArray =
                    str.replace("[", "").replace("]", "").split(",");
            System.out.println("numsArray: " + Arrays.toString(numsArray));

            int[] nums = new Solution().stringToInt(numsArray);
            System.out.println(new Solution().majorityElement(nums));
        } catch (Exception e) {
            // e.printStackTrace();
            // 回车一行就退出
            break;
        }
    }
}

 

c++

#include "iostream"
#include "vector"
using namespace std;

class Solution {
public:
    int majorityElement(vector<int>& nums) {
        int candidate = -1;
        int count = 0;
        for (int& num: nums) {
            if (count == 0) {
                candidate = num;
            }
            if (num == candidate) {
                count++;
            } else {
                count--;
            }
        }
        count = 0;
        int length = nums.size();
        for (int& num:nums) {
            if (num == candidate) {
                count++;
            }
        }
        return count * 2 > length ? candidate : -1;
    }
};

int main() {
    vector<int> nums = {1,2,5,9,5,9,5,5,5};
    cout << Solution().majorityElement(nums) << endl;
}

View Code

c

#include "stdio.h"

int majorityElement(int* nums, int numsSize) {
    int candidate = -1;
    int count = 0;
    for (int i=0; i<numsSize; i++) {
        if (count == 0) {
            candidate = nums[i];
        }
        if (nums[i] == candidate) {
            count++;
        } else {
            count--;
        }
    }
    count = 0;
    int length = numsSize;
    for (int i=0; i<numsSize; i++) {
        if (nums[i] == candidate) {
            count++;
        }
    }
    return count * 2 > length ? candidate : -1;
}

int main() {
    int nums[] = {1,2,5,9,5,9,5,5,5};
    int numsSize = sizeof(nums) / sizeof(nums[0]);
    printf("%d ", majorityElement(nums, numsSize));

}

View Code